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$F[X]$ is of the form $a_0+a_1X+...+a_nX^n$ where $a_0,a_1,...a_n\in F\text{ with characteristic p>0}.$ Express $f\in F[X]\ as\ g(X^{p^m})$, where the nonnegative integer m is a large as possible and $f$ is irreducible. Show that g is irreducible and separable.

This is a problem in section 3.4 of Basic Algebra by Robert Ash.

The question I had about is how to write $f(X)$ in form of $g(X^{p^m})$. I have tried this using the property $(a+b)^p=a^p+b^p$, since $F$ has characteristic p and the Frobenius Automorphism indicate if $\alpha \in F, \alpha = \beta ^p,$ for some $\beta \in F$. $g(X^{p^m})$ can be therefore written as $(b_0+b_1X+...+b_nX^n)^{p^m}.$ This is writing $g(X^{p^m})$ in in form of $f(X)$, for the other way around, is it just taking the $p^m$ root of the above expression: $$f(X)=\sqrt[p^m]{g(X^{p^m})}$$

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  • $\begingroup$ As you state this it is not true: for $p=2$, $f=X^4$, $g=X^2$ we have $f(X)=g(X^2)$ but $g$ is not irreducible. Perhaps you can type the question exactly it is in Ash? $\endgroup$ Feb 13, 2021 at 8:31
  • $\begingroup$ Thank you for reminding, I will change the question. $\endgroup$
    – John He
    Feb 13, 2021 at 8:34
  • $\begingroup$ But it is still false. Take $f=X(X+1)$, then we must have $g=f$ (with $m=0$) and $g$ is not irreducible. There must be more to the question than this. Is $f$ just possibly meant to be the minimal polynomial of some element in an extension field? $\endgroup$ Feb 13, 2021 at 9:21
  • $\begingroup$ I add the condition $f$ is irreducible $\endgroup$
    – John He
    Feb 13, 2021 at 19:37

1 Answer 1

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The fact that $g$ is irreducible follows from the fact that $f$ is; i.e. if $g = h_1 h_2$ then $f = h_1(x^{p^m})h_2(x^{p^m})$.

Once we show existence, this proves that $g$ is separable too: an irreducible polynomial is separable iff $g' \neq 0$, and in characteristic $p$ the fact that $g' = 0$ in turn means that $g = \widetilde{g}(x^p)$ for some $\widetilde g$. We cannot have the latter, otherwise it would contradict the maximality of $m$.

Now, consider

$S = \{m \in \mathbb N_0 : f = g(x^{p^m}), \text{ for some $g$}\}.$

The set $S$ is non empty, because $0 \in S$, and it is bounded above by degree considerations. Hence there exists $m = \max S$ with an associated $g$, which is irreducible and separable by the previous remarks.

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  • $\begingroup$ You mean inseparable iff g’=0? $\endgroup$
    – John He
    Feb 13, 2021 at 20:15
  • $\begingroup$ Yes, my bad. Fixed now! $\endgroup$
    – qualcuno
    Feb 13, 2021 at 20:16

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